If two circles (x - 1)^2 + (y - 3)^2 = r^2 an | vedclass

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(A) The first circle is $(x - 1)^2 + (y - 3)^2 = r^2$,with center $C_1 = (1, 3)$ and radius $r_1 = r$.The second circle is $x^2 + y^2 - 8x + 2y + 8 =

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$2 < r < 8$

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(A) The first circle is $(x - 1)^2 + (y - 3)^2 = r^2$,with center $C_1 = (1, 3)$ and radius $r_1 = r$.The second circle is $x^2 + y^2 - 8x + 2y + 8 = 0$. Rewriting in standard form: $(x^2 - 8x + 16) + (y^2 + 2y + 1) = -8 + 16 + 1$,which is $(x - 4)^2 + (y + 1)^2 = 9 = 3^2$. Thus,center $C_2 = (4, -1)$ and radius $r_2 = 3$.The distance between the centers $d = \sqrt{(4 - 1)^2 + (-1 - 3)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$.For two circles to intersect at two distinct points,the condition is $|r_1 - r_2| $1$) $r_1 + r_2 > d$ $\Rightarrow r + 3 > 5$ $\Rightarrow r > 2$.$2$) $|r_1 - r_2| Combining $r > 2$ and $r < 8$,we get $2 < r < 8$.

If two circles $(x - 1)^2 + (y - 3)^2 = r^2$ and $x^2 + y^2 - 8x + 2y + 8 = 0$ intersect in two distinct points,then

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